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Kazeer [188]
3 years ago
10

A sleepy student drops a calculator out of a window that's 20.7\text{ m}20.7 m20, point, 7, start text, space, m, end text off t

he ground. We can ignore air resistance.
What is the velocity of the calculator after falling for 1.8\,\text s1.8s1, point, 8, start text, s, end text?
Physics
1 answer:
Ira Lisetskai [31]3 years ago
8 0

Answer:

Explanation:

You can ignore how high the window actually is, unless the calculator falls further.  You just want to use the formula v_f = v_i + at  Let em know if that doesn't help and I can work you through it.

To check the distance use y = v_i*t + .5at^2.  Would just plug in 0*1.8 + .5(-9.8)1.8^2 = -15.876, so it doesn't fall more than 20.7 m so indeed, you don't have to worry about that.  

Again, let me know if you need further help.

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Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia
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Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass M_{Io}=8.93\times10^{22}\ kg

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Height h_{Io}=500\ km

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

g =\dfrac{GM_{Io}}{(R_{Io})^2}

Put the value into the formula

g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}

g=1.79\ m/s^2

Let  v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

H=\dfrac{v^2}{2g}

Using ratio of height of earth and height of Io

\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}

\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}

Put the value into the formula

\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}

\dfrac{H_{e}}{H_{Io}}=0.182

H_{e}=0.182\times H_{Io}

H_{e}=0.182\times500

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Hence, The height reached by the material on Earth is 91 km.

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The diver has the initial velocity, both (a) magnitude is 9.8 m/s and (b) direction is  73.5°.

<h3>What is free falling?</h3>

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If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E)

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

K.E = 1/2mv²

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A diver springs upward from a board that is 4.40 m above the water. At the instant, she contacts the water her speed is 13.5 m/s and her body makes an angle of 78.1 ° with respect to the horizontal surface of the water.

(a)

From energy conservation principle, initial and final mechanical energy are equal.

1/2mu² + mgh = 1/2mv²

where, u is the initial velocity of the diver.

u = sq rt  (v² - 2gh)

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Thus, the velocity of the diver is 9.8 m/s.

(b)

The horizontal component of velocity will remain constant.

The horizontal component of acceleration is zero.

Then,

ucosθ = vcosΦ

θ = cos⁻¹ [ (13.5 x cos 78.1)/9.8 ]

θ = 73.5°

Thus, the direction of velocity is  73.5°.

Learn more about free falling.

brainly.com/question/13299152

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